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Q)

On putting $y=vx,$ the homogeneous differential equation $x^{2}dy+y(x+y)dx=0$ becomes

\[\begin{array}{1 1}(1)xdv+(2v+v^{2})dx=0&(2)vdx+(2x+x^{2})dv=0\\(3)v^{2}dx-(x+x^{2})dv=0&(4)vdv+(2x+x^{2})dx=0\end{array}\]

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A)
$y= vx => \large\frac{dy}{dx} $$=v+x \large\frac{dv}{dx}$
$x^2dy= y(x+y) dx=0$
$x^2 dy =-y (x+y) dx$
$\large\frac{dy}{dx} =-\large\frac{y (x+y)}{x^2}$
$v +x $$\large\frac{dv}{dx} =\frac{-vx(x+vx)}{x^2}$
$x $$ \large\frac{dv}{dx} =\frac{-vx^2(1+v)}{x^2} $$-v$
$x \large\frac{dv}{dx} $$=-v -v^2-v$
$x \large\frac{dv}{dx} $$ =-v^2 -2v$
$x dv= -(v^2-2v)$
$xdv +(v^2 +2v)dx=0$
Hence 1 is the correct answer.
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